Question

Find a closed form expression for the nth right Riemann sum of this integral?

Answer 1

Partitioning the interval $[6,11]$ into $n$ equally-spaced subintervals gives $n$ rectangles of width $\Delta x=\dfrac{11-6}n=\dfrac5n$ and of heights determined by the right endpoints of each subinterval.

If $a=x_1=6$, then $x_2=6+\dfrac5n$, $x_3=6+2\dfrac5n$, and so on, up to $b=x_{n+1}=6+n\dfrac5n=11$. Because we're using the right endpoints, the approximation will consider $x_2,\ldots,x_{n+1}$

The definite integral is then approximated by

$\displaystyle\int_6^{11}(1-5x)\,\mathrm dx\approx\sum_{i=2}^{n+1}f(x_i)\Delta x=\sum_{i=2}^{n+1}\left(1-\left(6+(i-1)\dfrac5n\right)\dfrac5n$

You have

$\displaystyle\sum_{i=2}^{n+1}\left(1-5\left(6+(i-1)\dfrac5n\right)\dfrac5n=\sum_{i=2}^{n+1}\left(-29-\dfrac{25}n(i-1)\right)\dfrac5n$
$=\displaystyle-{145}n\sum_{i=2}^{n+1}1-\dfrac{125}{n^2}\sum_{i=2}^{n+1}(i-1)$
$=-145-\dfrac{125(n^2+n)}{2n^2}$
$=-\dfrac{145}2-\dfrac{125}{2n}$

To check that this is correct, let's make sure the sum converges to the exact value of the definite integral. As $n\to\infty$, you have the sum converging to $-\dfrac{145}2$.

Meanwhile,

$\int_6^{11}(1-5x)\,\mathrm dx=\left[x-\dfrac52x^2\right]_{x=6}^{x=11}=-\dfrac{415}2$

so we're done.